C++ from char array to BCD

Question

Arduino project using the 'Arduino language' (C++ wrapper) -

I have an 8-char array which gives 24Hr clock in the form {'H','H',':','M','M',':','S,'S'} and I need to convert this to BCD. I thought I understood the idea well enough but I get an unexpected result.

As a base example I'd like to convert {'1','2'} to BCD - the theory is on the left and my code result is on the right of the image below. I suspect the issue is with buff[0] being a 'number char', I tried atoi(buff[0]) but this still failed (I had the same result):

How may I do this correctly? My baseline code is:

char buff[8] = {'1','2',':','3','4',':','5','5'};    
uint8_t HourTens  = buff[0] && 0b11110000; // && 0xF0, should be 1
uint8_t HourUnits = buff[1] && 0b00001111; // && 0x0F, should be 2
uint8_t HourBCD   = HourTens + HourUnits;

Serial.println( buff );
Serial.print("'"); Serial.print( buff[0] ); Serial.print( "', HourTens:  " );   Serial.println( HourTens );
Serial.print("'"); Serial.print( buff[1] ); Serial.print( "', HourUnits: " ); Serial.println( HourUnits );
Serial.print( "HourBCD: " ); Serial.println( HourBCD );
Serial.println("***");

Answer 1

The problem with your code is that the data in the buff is ASCII characters, not the individual binary digits. You need to subtract '0' from each digit before performing a conversion to BCD:

uint8_t HourTens  = (buff[0] - '0'); // && 0xF0, will be 1

Note that each element takes its own byte, so you don't need to mask anything.

uint8_t HourBCD = ((buff[0] - '0') << 4) | (buff[1] - '0');

Also note that since the digits are from zero to nine, all you need to do is to shift the upper digit by four to get a valid BCD representation.